An electron has a spin quantum number +1/2 an | vedclass

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(C) For a given azimuthal quantum number $l$,the magnetic quantum number $m$ can take values from $-l$ to $+l$.For $s-$ orbital,$l = 0$,so $m$ can onl

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$s-$ orbital

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(C) For a given azimuthal quantum number $l$,the magnetic quantum number $m$ can take values from $-l$ to $+l$.For $s-$ orbital,$l = 0$,so $m$ can only be $0$.For $p-$ orbital,$l = 1$,so $m$ can be $-1, 0, +1$.For $d-$ orbital,$l = 2$,so $m$ can be $-2, -1, 0, +1, +2$.For $f-$ orbital,$l = 3$,so $m$ can be $-3, -2, -1, 0, +1, +2, +3$.Since the electron has $m = -1$,it can exist in $p, d,$ or $f$ orbitals,but it cannot be present in an $s-$ orbital because the only possible value for $m$ in an $s-$ orbital is $0$.

Column-$I$	Column-$II$
$(a)$ $4s$	$(p)$ Circular orbit around nucleus
$(b)$ $4p$	$(q)$ Non-directional orbital
$(c)$ $1s$	$(r)$ Angular momentum $= 2h/\pi$
$(d)$ $3d$	$(s)$ Radial node is zero

An electron has a spin quantum number $+1/2$ and magnetic quantum number $-1.$ It cannot be present in :

Similar Questions

Match the columns and choose the correct option:

Column-$I$ Column-$II$

$(a)$ $4s$ $(p)$ Circular orbit around nucleus

$(b)$ $4p$ $(q)$ Non-directional orbital

$(c)$ $1s$ $(r)$ Angular momentum $= 2h/\pi$

$(d)$ $3d$ $(s)$ Radial node is zero

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